How do you evaluate # e^( (3 pi)/2 i) - e^( ( 11 pi)/12 i)# using trigonometric functions?

Question

How do you evaluate #     e^( (3    pi)/2    i)  -   e^(  (  11   pi)/12     i)# using trigonometric functions?

Emery Adkins · Accepted Answer

#e^((3pi)/2i)- e^((11pi)/12i) ~~ 0.966 - 1.259 i# We know #e^(itheta) = cos theta +i sin theta# #(3pi)/2 =(3*180)/2= 270^0 , (11 pi)/12= (11*180)/12 = 165^0# #cos 270 =0 ; sin(270) = -1 ; cos 165 ~~ -0.966 ; # #sin 165 ~~ 0.259 # #:. e^((3pi)/2i) = cos ((3pi)/2)+ i sin ((3pi)/2) = 0 - i*1=0-i # #:. e^((11pi)/12i) = cos ((11pi)/12)+ i sin ((11pi)/12) ~~ -0.966 +0.259 i# # :. e^((3pi)/2i) - e^((11pi)/12i) ~~ (0- i) - (- 0.966 + 0.259 i)# or # ~~ (0.966) + (-1-0.259)i ~~ 0.966 - 1.259 i#

Daniel Acosta · Answer

undefined To evaluate $e^{\left(\frac{3\pi}{2}i\right)} - e^{\left(\frac{11\pi}{12}i\right)}$ using trigonometric functions, we can express each exponential term in trigonometric form.

Recall Euler's formula: $e^{ix} = \cos(x) + i\sin(x)$.

So,

$e^{\left(\frac{3\pi}{2}i\right)} = \cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right)$.

$\cos\left(\frac{3\pi}{2}\right) = 0$ and $\sin\left(\frac{3\pi}{2}\right) = -1$, so $e^{\left(\frac{3\pi}{2}i\right)} = -i$.

Similarly,

$e^{\left(\frac{11\pi}{12}i\right)} = \cos\left(\frac{11\pi}{12}\right) + i\sin\left(\frac{11\pi}{12}\right)$.

Using the half-angle identities, $\cos\left(\frac{11\pi}{12}\right)$ and $\sin\left(\frac{11\pi}{12}\right)$ can be computed.

Now, we can substitute these values back into the expression $e^{\left(\frac{3\pi}{2}i\right)} - e^{\left(\frac{11\pi}{12}i\right)}$:

$(-i) - (\cos\left(\frac{11\pi}{12}\right) + i\sin\left(\frac{11\pi}{12}\right))$.

Combining the real and imaginary parts, we get:

$(-i) - \cos\left(\frac{11\pi}{12}\right) - i\sin\left(\frac{11\pi}{12}\right)$.

Re-arranging terms, we have:

$-\cos\left(\frac{11\pi}{12}\right) - i\sin\left(\frac{11\pi}{12}\right) - i$.

This is the final answer.